MHB Ryan's question at Yahoo Answers (Eigenvalues of A^*A)

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All eigenvalues of the matrix A*A are proven to be nonnegative by showing that for any eigenvalue λ, it can be expressed as the ratio of the squared norm of Ax to the squared norm of x, which is always nonnegative. Additionally, the matrix A*A + I is shown to be invertible by demonstrating that if -1 were an eigenvalue of A*A, it would lead to a contradiction regarding the determinant. Thus, the determinant of A*A + I is non-zero, confirming its invertibility. This discussion provides a clear understanding of these linear algebra concepts. Further inquiries can be directed to the specified math help forum.
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Here is the question:

Let A be a mxn matrix. Prove that all eigenvalues of A*A are nonnegative and prove that A*A + I is invertible.

THANK YOU!

Here is a link to the question:

Linear Algebra Proof Help Please? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello Ryan,

Suppose $\lambda$ is an eigenvalue of $A^*A\in \mathbb{C}^{n\times n}$ then, there exists $x=(x_1,\ldots,x_n)^T\in\mathbb{C}^n$ such that $A^*Ax=\lambda x$. Multypling both sides by $x^*=(\overline{x_1},\ldots,\overline{x_n})$ we get $$x^*A^*Ax=(Ax)^*(Ax)=\lambda x^*x$$ But $y=Ax=(y_1,\ldots,y_m)^T$ and $x\neq 0$ (i.e. $||x||\neq0$), so $$\lambda=\frac{y^*y}{x^*x}=\frac{||y||^2}{||x||^2}\ge 0$$ On the other hand, $$\det (A^*A+I)=0\Leftrightarrow \det (A^*A-(-1)I)=0\Leftrightarrow -1\mbox{ is eigenvalue of }A^*A$$ This is a contradiction, so $\det(A^*A+I)\neq0$, as a consequence $A^*A+I$ is invertible.

If you have further questions, you can post them in the http://www.mathhelpboards.com/f14/ section.
 
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